高中排列组合问题。求解答。

The answer is 144.

First, we have 3! ways to fulfil the first row.
Seoncd, we have 3! ways to fulfil the first column.
Now, we can suppose, without loss of generality,
that the first row is filled by 1,2,3,4 from left to right,
and the first column is filled by 1,2,3,4 from top to bottom.

Next, to fill the entry at Row 2 and Column 2. It is clear that this entry
can be filled by 1 or 3 or 4. So we have three cases:

Case 1. The entry at Row 2 and Column 2 is filled by the number 1.
In this case, the second row must be 2,1,4,3 from left to right,
and the second column must be 2,1,4,3 from top to bottom.
The remaining 2*2 squares will be filled by two 1's and two 2's.
There is two ways to do that, obviously.
Therefore, we have two ways to fulfil the 4*4 square in total.

Case 2. The entry at Row 2 and Column 2 is filled by the number 3.
In this case, the second row must be 2,3,4,1 from left to right,
and the second column must be 2,3,4,1 from top to bottom.
Consequently, the third row must be 3,4,1,2 from left to right,
and the fourth row must be 4,1,2,3 from left to right.
Therefore, we have only one way to fulfil the 4*4 square in total.

Case 3. The entry at Row 2 and Column 2 is filled by the number 4.
Observe that the numbers 3 and 4 are symmetric in the given configuration.
So the number of ways must be the same as in Case 2, that is, one way.

Finally, we add them up, and find the total number of fulfilling
the square is 3!^2*(2+1*2)=144. This completes the solution.